Find the smallest number which when divided by 48,56,105 and 225 leaves the same remainder 7 in each case
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LCM of 48, 56, 105, 225 = 25200
25200 id the smallest no. divisible by 48, 56, 105, 225 So the remainder is 0. But if we want the remainder to be 7, then we add 7 to the no.
Hence the required the no. 25207
The ans is 25207
25200 id the smallest no. divisible by 48, 56, 105, 225 So the remainder is 0. But if we want the remainder to be 7, then we add 7 to the no.
Hence the required the no. 25207
The ans is 25207
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Step-by-step explanation:
फाइंड आउट द स्मालेस्ट नंबर दैट कैन वी डिवाइड बाय 48and98 एंड 105 इन ए रिमाइंडर 7
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